A ship leaves base and sails 30 km on a bearing of 034. It is then ordered to take up position due North of the base in exactly 1 hour's time. If the ship has a speed of 20km/h, find the two possible bearings it could take.

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- May 16th 2009, 08:43 PMwomatamaHow do you solve this bearing trig problem?
A ship leaves base and sails 30 km on a bearing of 034. It is then ordered to take up position due North of the base in exactly 1 hour's time. If the ship has a speed of 20km/h, find the two possible bearings it could take.

- May 16th 2009, 10:34 PMGrandad
Hello womatama

Welcome to Math Help Forum!In the attached diagram, the ship's two possible courses are given by $\displaystyle BC_1$ and $\displaystyle BC_2$. Then use the Sine Rule:

$\displaystyle \frac{\sin\theta}{30}=\frac{\sin34^o}{20}$

$\displaystyle \Rightarrow \sin\theta =0.8388$

There are then two possible values of $\displaystyle \theta: \theta_2 = 57.0^o$ and $\displaystyle \theta_1 = 180 - 57.0 = 123.0^o$

So the two possible courses to steer are $\displaystyle 360 - \theta_2 = 303^o$ and $\displaystyle 360-\theta_1 = 237^o$.

Grandad - May 16th 2009, 11:43 PMwomatama
- May 17th 2009, 12:17 AMGrandad
Hello womatamaI don't know how much you know about sines and cosines, but amng the rules are, for any value of $\displaystyle \theta$:

- $\displaystyle \sin\theta = \sin(180-\theta)$

- $\displaystyle \cos\theta = \cos(180-\theta)$

which tells you how to find the sine and cosine of obtuse angles. So, for example, $\displaystyle \sin57^o = \sin(180-57)^o=\sin123^o$ (and $\displaystyle \cos 123^o = -\cos57^o$). Check these out on your calculator if you like.

Grandad

- May 17th 2009, 01:26 AMwomatama
Oh i get it now

i looked at the sine graph and cosine graph and then i understood it. THX